Q70 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=$

  • A
    $\frac{1}{10}$
  • Correct answer
    B
    $-\frac{1}{10}$
  • C
    $\frac{2}{5}$
  • D
    $-\frac{2}{5}$

Solution

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$ $\begin{aligned} & =\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(2 x+3)(x-1)} \times \frac{(\sqrt{x}+1)}{(\sqrt{x}+1)} \\ & =\lim _{x \rightarrow 1} \frac{(2 x-3)(x-1)}{(2 x+3)(x-1)(\sqrt{x}+1)} \\ & =\lim _{x \rightarrow 1} \frac{(2 x-3)}{(2 x+3)(\sqrt{x}+1)}=\frac{-1}{5(2)}=\frac{-1}{10} .\end{aligned}$

Hence, the correct answer is option: 'B'.

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